TEST BANK FOR Adaptive Filter Theory 4th Edition By Simon Haykin

1.1 Let
(1)
(2)
We are given that
(3)
Hence, substituting Eq. (3) into (2), and then using Eq. (1), we get
1.2 We know that the correlation matrix R is Hermitian; that is
Given that the inverse matrix R-1 exists, we may write
where I is the identity matrix. Taking the Hermitian transpose of both sides:
Hence,
That is, the inverse matrix R-1 is Hermitian.
1.3 For the case of a two-by-two matrix, we may
ru(k) = E[u(n)u*(n – k)]
ry(k) = E[y(n)y*(n – k)]
y(n) = u(n + a) – u(n – a)
ry(k) = E[(u(n + a) – u(n – a))(u*(n + a – k) – u*(n – a – k))]
= 2ru(k) – ru(2a + k) – ru(– 2a + k)
RH = R
R –1RH = I
RR –H = I
R –H R –1 =
Ru = Rs + Rν
2
For Ru to be nonsingular, we require
With r12 = r21 for real data, this condition reduces to
Since this is quadratic in , we may impose the following condition on for nonsingularity
of Ru:
where
1.4 We are given
This matrix is positive definite because
r11 r12
r21 r22
σ2 0
0 σ2
= +
r11 σ2 + r12
r21 r22 σ2 +
=
det(Ru) r11 σ2 ( + ) r22 σ2 = ( + ) – r12r21 > 0
r11 σ2 ( + ) r22 σ2 ( + ) – r12r21 > 0
σ2 σ2
σ2 1
2
--(r11 + r22) 1
4Δr
(r11 + r22)2 – 1
– --------------------------------------
 
 
 
>
Δr r11r22 r12
2 = –
R 1 1
1 1
=
aTRa [a1,a2] 1 1
1 1
a1
a2
=
a1
2 2a1a2 a2
2 = + +
3
for all nonzero values of a1 and a2
(Positive definiteness is stronger than nonnegative definiteness.)
But the matrix R is singular because
Hence, it is possible for a matrix to be positive definite and yet it can be singular.
1.5 (a)
(1)
Let
(2)
where a, b and C are to be determined. Multiplying (1) by (2):
where IM+1 is the identity matrix. Therefore,
(3)
(4)
(5)
(6)
From Eq. (4):
(a1 + a2)2 = > 0
det(R) (1)2 (1)2 = – = 0
RM+1
r(0)
r
rH
RM
=
RM+1
–1 a
b
bH
C
=
IM+1
r(0)
r
rH
RM
a
b
bH
C
=
r(0)a rH+ b = 1
ra + RMb = 0
rbH + RMC = IM
r(0)bH rH+ C 0T =
4
(7)
Hence, from (3) and (7):
(8)
Correspondingly,
(9)
From (5):
(10)
As a check, the results of Eqs. (9) and (10) should satisfy Eq. (6).
We have thus shown that
b RM
–1= – ra
a
1
r(0) rHRM
–1– r
= ------------------------------------
b
RM
–1r
r(0) rHRM
–1– r
= – ------------------------------------
C RM
–1 RM
–1rbH = –
RM
–1 RM
–1rrHRM
–1
r(0) rHRM
–1– r
= + ------------------------------------
r(0)bH rH+ C
r(0)rHRM
–1
r(0) rHRM
–1– r
– ------------------------------------ rHRM
–1 rHRM
–1rrHRM
–1
r(0) rHRM
–1– r
= + + ------------------------------------
0T =
RM+1
–1 0
0
0T
RM
–1 a
1
RM
–1r
rHRM
–1 –
RM
–1rrHRM
–1
= +
0
0
0T
RM
–1 a
1
RM
–1– r
1 rHRM
–1 = + [ – ]
5
where the scalar a is defined by Eq. (8):
(b) (11)
Let
(12)
where D, e and f are to be determined. Multiplying (11) by (12):
Therefore
(13)
(14)
(15)
(16)
From (14):
(17)
Hence, from (15) and (17):
(18)
Correspondingly,
RM+1
RM
rBT
rB*
r(0)
=
RM+1
–1 D
eH
e
f
=
IM+1
RM
rBT
rB*
r(0)
=
D
eH
e
f
RMD rB*eH + = I
RMe rB* + f = 0
rBTe + r(0) f = 1
rBTD r(0)eH + 0T =
e – RM
–1rB* = f
f
1
r(0) rBTRM
–1rB* –
= ---------------------------------------------
6
(19)
From (13):
(20)
As a check, the results of Eqs. (19) and (20) must satisfy Eq. (16). Thus
We have thus shown that
where the scalar f is defined by Eq. (18).
1.6 (a) We express the difference equation describing the first-order AR process u(n) as
where w1 = -a1. Solving this equation by repeated substitution, we get
e = –
RM
–1rB*
r(0) rBTRM
–1rB* –
---------------------------------------------
D RM
–1 RM
–1rB*eH = –
RM
–1 RM
–1rB*rBTRM
–1
r(0) rBTRM
–1rB* –
= + ---------------------------------------------
rBTD r(0)eH + rBTRM
–1 rBTRM
–1rB*rBTRM
–1
r(0) rBTRM
–1rB* –
------------------------------------------------
r(0)rBTRM
–1
r(0) rBTRM
–1rB* –
= + – ---------------------------------------------
0T =
RM+1
–1 RM
–1
0T
0
0
f
RM
–1rB*rBTRM
–1
rBTRM
–1 –
RM
–1rB* –
1
= +
RM
–1
0T
0
0
f
–RM
–1rB*
1
= + rBT RM
–1 [– 1]
u(n) = v(n) + w1u(n – 1)
u(n) = v(n) + w1v(n – 1) + w1u(n – 2)